KROHNE MFC 010 C Converter - Manual (Page 95)

MFC010 Interface Manual 93
Appendix C : Decoding Floating Point Numbers
Floating point numbers in computer s ystems ar e us ed t o r epresent v alues ov er a l arger r ange t han i s
practical with standard integers. The representation of the value is achieved by splitting it into three parts,
the sign “S” (±), the Exponent “E” and the Mantissa or Fraction “M”.
Single Precision Floating Point Numbers (“Floats”)
Single precision floating-point values are coded in a gr oup of 4 by tes ( 32 bi ts, as shown below). T his
enables the computer to represent values over a range of ±3.4 x 10 ±38 , with an accuracy of 7 decimal
digits.
S E E E E E E E E M M M M M M M M M M M M M M M M M M M M M M M
Byte 3 (MSB) Byte 2 Byte 1 Byte 0 (LSB)
To encode a value into a floating-point representation, use the following process.
Step 1 : If Value < 0 (i.e. negative) S = 1, otherwise S = 0
Step 2 : Set E = 127
Step 3 : If Value < 2, skip to step 6
Step 4 : Divide the Value by 2, add 1 to E
Step 5 : Go back to step 3
Step 6 : If Value > 1, skip to step 9
Step 7 : Multiply the Value by 2, subtract 1 from E
Step 8 : Go back to step 6
Step 9 : M = ( Value – 1 ) * 2 23
For Example :
Value = 206171.125
Step 1 : Value > 0, so S = 0
Step 2 : E = 127
Step 3 – Step 6, Divide value by 2, 17 times, E = 127 + 17 = 144 = 90
16
Value = 206171.125 / 2 17 = 1.5729609
Step 9 : M = ( 1.5729609 – 1 ) * 2 23 = 4806344 = 4956C8
16
Therefore the result is 0100 1000 0100 1001 0101 0110 1100 1000 = 484956C8
16
To decode a floating point representation use the following formula.
Floating Point Value = -1 S x 2 (E-127) x [ 1 + { M / 2 23 } ]
Working backwards from the previous example :
Floating point Number 484956C8
16
= 0100 1000 0100 1001 0101 0110 1100 1000
2
So….. S = 0
E = 100 1000 0 = 90
16
= 144
M = 0100 1001 0101 0110 1100 1000 = 4956C8
16
= 4806344